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Surface integral of vector field

  1. Jul 13, 2014 #1
    1. The problem statement, all variables and given/known data
    find ∫E.dS, where E = (Ar^2, Br (sinθ),C cosρ), over the outside conical surface S, given by 1≤r≤2, θ=[itex]\pi[/itex]/3 (this is an open surface, excluding the end faces).


    2. Relevant equations



    3. The attempt at a solution

    from the context I believe ρ is the plane polar angle on the x-y plane, so the surface is a slanted ring with width 1 around the z axis.
    expressing S in vector v = ((√3 r cosρ)/2, (√3 r sinρ)/2, r/2), so ∫E.dS = ∫[itex]^{2pi}_{0}[/itex]∫[itex]^{2}_{1}[/itex] (Ar^2, (√3 Br)/2 ,C cosρ).((-√3 r cosρ)/4, (-√3 r sinρ)/4, 3r/4) drdρ, where I have taken the normal to the surface from outside to inside of the conical. Then, after the dot product each term in the integrand has a sine or cosine term, hence integrating ρ from 0 to 2pi will give zero.

    the given answer is (7B pi)/2. where did I go wrong?
     
    Last edited: Jul 13, 2014
  2. jcsd
  3. Jul 13, 2014 #2
    From the problem statement I would rather say (r, ##\theta##, ##\rho##) are spherical coordinates.
     
  4. Jul 13, 2014 #3
    yes thats what I've used. using let θ=pi/3 I parameterise the surface in terms of ρ and r. did I make any mistake here?
     
  5. Jul 13, 2014 #4
    The domain is right: 1<r<2 and 0<##\rho##<2 pi, but the vectors are expressed in the basis ##\{e_{r},e_{\theta},e_{\rho}\}##.
    The unit normal outside the surface is then very simple.
    Can you find dS in this basis?
     
    Last edited: Jul 13, 2014
  6. Jul 13, 2014 #5
    hmm I dont think I get it. partial differentiating v wrt r and ρ, then take the cross product of the 2, I got ((-√3 r cosρ)/4, (-√3 r sinρ)/4, 3r/4) , pointing into the ring. is this not the correct normal?
     
  7. Jul 13, 2014 #6
    Ok, yes it is possible but way too complicate here. (E is expressed in the basis I mentioned above. You are expressing things in ##\{e_x,e_y,e_z\}## here so to take the dot product you would have to express E in ##\{e_x,e_y,e_z\}##)
    Keep things in shperical coordinates, it is much simpler here.
    Remember
    ##e_{r}##: unite vector in the direction of growing r
    ##e_{\theta}##: unite vector in the direction of growing ##\theta##
    ##e_{\rho}##: unite vector in the direction of growing ##\rho##
     
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